Graded Rings

This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.

Graded Rings

When the index set \(I\) is a commutative monoid, we agreed to call the direct sum \(\bigoplus_{i\in I} A_i\) of a family of abelian groups \((A_i)_{i\in I}\) a graded abelian group. At that time there was no additional condition on the \(A_i\), so this was not a particularly interesting definition; but now that a multiplicative structure is imposed on the \(A_i\), the definition becomes more meaningful.

Definition 1 Let \(I\) be a commutative monoid and let \((A_i)_{i\in I}\) be an \(I\)-indexed family of abelian groups. If the multiplication defined on \(\bigoplus_{i\in I} A_i\) makes it into a ring and, in addition, the condition

\[A_i A_j\subseteq A_{i+j}\qquad\text{for all $i,j\in I$}\]

is satisfied, then we call \(A\) a graded ring indexed by \(I\). An element belonging to \(A_i\) is called a homogeneous element.

By definition, any element of \(A\) can be written uniquely as a finite sum of homogeneous elements.

Proposition 2 Suppose every element of \(I\) is cancellable and that \(\bigoplus_{i\in I} A_i\) is a graded ring. Then \(A_0\) is a subring of \(A\).

Proof

From \(A_0A_0\subseteq A_0\) it is obvious that \(A_0\) is closed under multiplication. Hence it suffices to show that the multiplicative identity \(1\) of \(A=\bigoplus A_i\) lies in \(A_0\). Write \(1=\sum_{i\in I} \epsilon_i\). Then for any \(\alpha\in A_j\),

\[\alpha=1\alpha=\sum_{i\in I} \epsilon_i\alpha\in A_j\]

and therefore \(\epsilon_i\alpha=0\) for all \(i\neq 0\), while \(\epsilon_0\alpha=\alpha\) holds only for \(i=0\). Since any element of \(A\) can be expressed as a sum of homogeneous elements, the proof is complete.

In most cases of interest we have either \(I=\mathbb{Z}\) or \(I= \mathbb{N}\). Hence the hypothesis of Proposition 2 is satisfied.

Example 3 For any abelian group \(G\), the free ring \(F(G)\) generated by \(G\) is an \(\mathbb{N}\)-graded ring.

Graded Ring Homomorphisms

Definition 4 Let \(I\) be a commutative monoid and let \(A,A'\) be two \(I\)-graded rings. A ring homomorphism \(\phi:A \rightarrow A'\) is called a graded homomorphism if \(\phi(A_i)\subseteq A_i'\) holds for every \(i\in I\).

It is not difficult to see that \(I\)-graded rings and \(I\)-graded homomorphisms form a category \(\bgr_I\Ring\).

Homogeneous Ideals and Quotients of Graded Rings

For a graded ring \(A=\bigoplus_{i\in I} A_i\), the quotient ring \(A/\mathfrak{a}\) of an ideal \(\mathfrak{a}\) of \(A\) need not always be a graded ring.

Example 5 Fix a ring \(A\) and consider the polynomial ring

\[A[\x]=\{\alpha_n\x^n+\cdots+\alpha_1\x+\alpha_0\mid \alpha_i\in A\}\]

whose coefficients are elements of \(A\). Then this ring carries a graded ring structure via the decomposition

\[A[\x]=\bigoplus_{n\geq 0} A\x^n.\]

On the other hand, consider the ideal \((\x-1)\) generated by \(\x-1\). Then as rings

\[A[\x]/(\x-1)\cong A\]

and explicitly this isomorphism is obtained by applying the first isomorphism theorem to the evaluation map defined by

\[\alpha_n\x^n +\cdots+\alpha_1\x+\alpha_0\quad \mapsto\quad \alpha_n+\cdots+\alpha_1+\alpha_0.\]

However, the above homomorphism is not a graded homomorphism.

To avoid this we introduce the notion of a homogeneous ideal.

Proposition 6 Let \(A=\bigoplus_{i\in I} A_i\) be an \(I\)-graded ring and let \(\mathfrak{a}\) be an ideal of \(A\). Then the following are all equivalent.

1. \(\mathfrak{a}\) is the sum of the \(\mathfrak{a}\cap A_i\).
2. Whenever an arbitrary element of \(\mathfrak{a}\) is decomposed into homogeneous elements, each of those homogeneous elements also lies in \(\mathfrak{a}\).
3. \(\mathfrak{a}\) is generated by homogeneous elements.

Proof

As elements of \(A\), all elements of \(\mathfrak{a}\) can be written uniquely as sums of homogeneous elements. Hence the equivalence of the first two conditions is obvious, and the first condition trivially implies the third. Now assume the third condition and prove the second. Suppose \(\mathfrak{a}\) is generated by homogeneous elements \((x_j)_{j\in J}\). Then an arbitrary \(x\in \mathfrak{a}\) can be written as

\[x=\sum_{j\in J} \alpha_j x_j,\qquad\text{$(\alpha_j)_{j\in J}$ finitely supported}.\]

Now each \(\alpha_j\) can in turn be written, as an element of \(A\), as a sum of homogeneous elements

\[\alpha_j=\sum_{k\in K_j} \alpha_{jk},\qquad \text{$(\alpha_{jk})_{k\in K_j}$ finitely supported}.\]

Hence

\[x=\sum_{j\in J}\sum_{k\in K_j}\alpha_{jk}x_j,\qquad \text{$(\alpha_{jk})_{j\in J,k\in K_j}$ finitely supported}\]

and the \(\alpha_{jk}x_j\) are each homogeneous elements and all belong to \(\mathfrak{a}\). From this the second condition follows.

An ideal satisfying the above equivalent conditions is called a homogeneous ideal. Then the following holds.

Proposition 7 For a homogeneous ideal \(\mathfrak{a}\), the quotient \(A/\mathfrak{a}\) is a graded ring and its decomposition is given by

\[A/\mathfrak{a}=\bigoplus_{i\in I}A_i/(\mathfrak{a}\cap A_i).\]

The proof of this is obvious and hence omitted.